Question

1. the x-intercepts of the parabola $y = x^2 - 10x + 23$ are given in exact form by

(1) $5 \pm \sqrt{2}$
(3) $4 \pm \sqrt{3}$
(2) $10 \pm \sqrt{2}$
(4) $7 \pm \sqrt{3}$

Explanation:

Step1: Set y=0 for x-intercepts

To find the x-intercepts of the parabola \( y = x^2 - 10x + 23 \), we set \( y = 0 \). So we have the quadratic equation:
\[
x^2 - 10x + 23 = 0
\]

Step2: Use quadratic formula

The quadratic formula for a quadratic equation \( ax^2 + bx + c = 0 \) is \( x=\frac{-b\pm\sqrt{b^2 - 4ac}}{2a} \). For the equation \( x^2 - 10x + 23 = 0 \), we have \( a = 1 \), \( b=- 10 \) and \( c = 23 \).
First, calculate the discriminant \( D=b^2-4ac \):
\[
D=(-10)^2-4\times1\times23=100 - 92 = 8
\]
Then, substitute \( a \), \( b \) and \( D \) into the quadratic formula:
\[
x=\frac{-(-10)\pm\sqrt{8}}{2\times1}=\frac{10\pm2\sqrt{2}}{2}=5\pm\sqrt{2}
\]

Answer:

(1) \( 5\pm\sqrt{2} \)